package algorithm.poj.p2000;

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.URL;
import java.net.URLDecoder;
import java.text.DecimalFormat;
import java.util.StringTokenizer;


/**
 * 分析：
 * 参考《100个著名初等数学问题》之“68 欧拉四面体问题”
 * 
 * 实现：
 * 
 * 经验和教训：
 * 精度
 * 
 * 分类：
 * 数学
 * 
 * 
 * @author wong.tong@gmail.com
 *
 */
public class P2208 {

	public static void main(String[] args) throws Exception {

		InputStream input = null;
		if (false) {
			input = System.in;
		} else {
			URL url = P2208.class.getResource("P2208.txt");
			File file = new File(URLDecoder.decode(url.getPath(), "UTF-8"));
			input = new FileInputStream(file);
		}
		
		BufferedReader stdin = new BufferedReader(new InputStreamReader(input));

		String line = stdin.readLine();
		StringTokenizer st = new StringTokenizer(line);
		int AB = Integer.valueOf(st.nextToken());
		int AC = Integer.valueOf(st.nextToken());
		int AD = Integer.valueOf(st.nextToken());
		int BC = Integer.valueOf(st.nextToken());
		int BD = Integer.valueOf(st.nextToken());
		int CD = Integer.valueOf(st.nextToken());

		int p = AB, q = AC, r = AD;
		int a = CD, b = BD, c = BC;
		
		long pp = p*p, qq = q*q, rr = r*r;
		long pq = pp+qq-c*c;
		long qr = qq+rr-a*a;
		long rp = rr+pp-b*b;
		long qp = pq, rq = qr, pr = rp;
		pp *= 2;
		qq *= 2;
		rr *= 2;
		
		long S = (pp*qq*rr + pr*rq*qp + pq*qr*rp) - (pp*qr*rq + qq*pr*rp + rr*pq*qp);	//S = 288*T^2
		//double T = Math.sqrt(S/288.0f);	//这一句居然WA
		double T = Math.sqrt(S)/Math.sqrt(288.0f);
		System.out.println(new DecimalFormat("##0.0000").format(T));
	}
}